Solve the following equation by factorisation:

$\dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6}$

Given,

$\dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \dfrac{1}{6} \\[1em] \Rightarrow x + 5 - x + 3 = \dfrac{(x - 3)(x + 5)}{6} \\[1em] \Rightarrow 8 = \dfrac{x^2 + 5x - 3x - 15 }{6} \\[1em] \Rightarrow 8 \times 6 = x^2 + 2x - 15 \\[1em] \Rightarrow 48 = x^2 + 2x - 15 \\[1em] \Rightarrow x^2 + 2x - 15 = 48 \\[1em] \Rightarrow x^2 + 2x - 15 - 48 = 0 \\[1em] \Rightarrow x^2 + 2x - 63 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 + 9x - 7x - 63 = 0 \\[1em] \Rightarrow x(x + 9) - 7(x + 9) = 0 \\[1em] \Rightarrow (x - 7)(x + 9) = 0 \text{ (Factorising left side) } \\[1em] x - 7 = 0 \text{ or } x + 9 = 0 \text{ (Zero-product rule) } \\[1em] x = 7 \text{ or } x = -9 \\[1em]$

Hence, the roots of given equation are -9 , 7.